[$]u=f\left(1+\frac{y-1}{1-t(y-1)}\right)[$]

Thanks. I'll get back on this.

I am working under the assumption that it is the solution of [$]\frac{\partial u}{\partial t} - \frac{\partial u}{\partial y} = 0 [$]

No, Paul is just posting my answer again for the problem you posted.

It is kind of interesting to see what happens there for initial data on [$]y > 1[$].

Suppose initial data [$]f(y)[$] over the whole positive y-axis.

Then, picture [$]u(y,t)[$] at fixed time-slice t for all [$]y>0[$].

As we already have seen, for y=0 to y=1 in the time-slice plot, we just see a somewhat distorted image of [$]f(y)[$] originally from [$]y=t/(t+1)[$] to [$]y=1[$], the fixed point. So, it is kind of like using a microscope to examine just part of the initial data (parts near [$]y=1[$]). As t gets larger, so does the power of the microscope.

For y>1, you see an image of

[$]f(y)[$] originally from [$]y=1[$] to [$]y=\infty[$], squished into the time-slice interval [$]y=1[$] to [$]y = y^*(t) \equiv 1 + 1/t[$]. So, this is the opposite idea: the whole range of the initial data is packed into a finite interval. Kind of like a trash compactor. As t gets larger, the size of the compactified trash gets smaller.

For [$]y > y^*(t)[$], I think we are starting to examine initial data on [$]y < 0[$], which wasn't defined by me (above), so probably best to say [$]u(y,t)[$] is undefined.